Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 13, Problem 49P
To determine

The mass of lead should be added to give the diver neutral buoyancy.

Expert Solution & Answer
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Answer to Problem 49P

  3.9kg

Explanation of Solution

Given:

Mass of a diver, m=85 kg

Density of a diver, ρ=0.96 kg/L

Formula used:

Let, volume of the diver be V

Density of the diver be ρD

Volume of the added lead be VPb

Mass of the lead be mPb

Let us apply Fy=0 to the diver,

  BWDWPb=0  (1)

Where, B is buoyant force

  B=ρwVD+PbgB=ρwVDg+ρwVPbg

  WD is weight of a diver

  WD=mDg=ρDVDg

  WPb is weight of the lead, WPb=mPbg

Substituting for B , WD , and WPb in equation (1) ,

  ρwVDg+ρwVPbgρDVDgmPbg=0

Or

  ρwVD+ρwVPbρDVDmPb=0  (2)

Where, ρw is density of water,

  VD is volume of a diver,

  g is acceleration due to gravity,

  VPb is volume of the lead,

  ρD is density of a diver,

  mPb is mass of a lead.

Mass (m) , density (ρ) , and volume (V) of a given substance is related through the expression,

  m=ρVρ=mV

Applying this to the equation (2) ,

  ρwmDρD+ρwmPbρPbρDmDρDmPb=0

  ρwmDρDmD+ρwmPbρPbmPb=0

  mD(ρwρD1)+mPb(ρwρPb1)=0

  mD(ρwρDρD)+mPb(ρwρPbρPb)=0

  mPb(ρwρPbρPb)=mD(ρwρDρD)

or

  mPb(ρPbρwρPb)=mD(ρwρDρD)

or

  mPb(ρPbρwρPb)=mD(ρwρDρD)

  mPb=mD(ρwρDρD)(ρPbρwρPb)=ρPb(ρwρD)mDρD(ρPbρw)  (3)

Calculation:

Density of lead, ρPb=11.3×103kg/m3

Density of water, ρw=1.0×103kg/m3

Substituting the numerical values in equation (3) ,

  mPb=(11.3×103kg/m3)(1×103kg/m3-0.96×103kg/m3)85 kg(0.96×103kg/m3)(11.3×103kg/m3-1×103kg/m3)=3.9kg

Conclusion:

The mass of lead should be added to give the diver neutral buoyancy 3.9kg .

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